{"id":"ee466a88-55c5-4a42-bc1b-a1b20d01a2c9","arxiv_id":"1908.05456","paper_version":2,"verdict":"ACCEPT","confidence":"MODERATE","novelty_score":6.0,"correctness_risk":"low","formal_verification":"none","parameter_count":0,"one_line_summary":"For two-element generating sets, infinitely many pairs (C_m*C_n, PSL(2,q)) fail the Neumanns' lifting property, and the first infinite quotients with non-liftable generating pairs are constructed.","lead":"This paper proves that many groups built from free products of cyclic groups, such as C2*C3, can map onto projective special linear groups PSL(2,q) in a way where some generating pairs of the image cannot be pulled back to generating pairs of the source. It gives infinitely many such counterexamples and the first ones where the image group is infinite.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"No significant objection identified","rationale":"The paper's central claim is an existence theorem about non-lifting generating pairs. The reduction in Corollary 2.3 is sound: the only way a generating pair of H can lift to G = Cm*Cn is if its Aut(F2)-orbit contains an (m,n)-generating pair. Fact 2.2 supplies the required transitivity of the Aut(F2)-action on generating pairs of Cm*Cn, and the cited Grushko–Neumann argument is standard and correct. The trace invariant is well-defined and constant on orbits, so the strategy of finding an orbit whose trace is not attained by any (m,n)-generating pair is valid. The case analysis in Theorem 1.1 follows this strategy consistently. In part (i), the two branches (trace 0, and trace ±1) correctly reduce to Miller's classification of groups generated by elements of orders 2 and 3 with commutator of order 2 or 3; the groups PSL(2,q) in question are non-solvable, so the corresponding orbits are (2,3)-free. Part (ii) computes the commutator trace for a (2,p)-generating pair as s^2+2 and then uses Theorem 2.4 to choose an orbit outside that set. The apparent statement that the entries of B lie in F_p rather than F_q is the only notable slip, but the subsequent use of s in F_q and the algebraic identities over F_q make clear that the intended proof works for all q = p^k. Parts (iii) and (iv) are direct applications of Corollary 2.7 and the finite lemmas, and the finite checks in Lemmas 3.1 and 3.2, while compressed, cover the required cases. Corollary 1.2's construction of an infinite H is also sound: the diagonal map is surjective because H0 is perfect and G/G'' is metabelian. Overall, the argument is coherent and the central claim is supported. I do not see a load-bearing weakness that would change the reader's ACCEPT verdict.","tokens_in":11172,"tokens_out":42279,"duration_ms":414668,"concrete_test":"In GAP, verify for a prime power q = p^k with k > 1, e.g. q = 25, that every generating pair of PSL(2,q) consisting of an involution and an element of order p has commutator trace of the form s^2+2 for some s in F_q; this confirms that the F_p/F_q typo in the proof of Theorem 1.1(ii) is not load-bearing.","verdict_should_be":"UNCHANGED","load_bearing_attack":"After checking the reduction via Corollary 2.3, the use of Fact 2.2, the trace-invariant computations, and the finite-exception cases, I do not find a load-bearing error. Fact 2.2 is the key external premise, and it is correct: Grushko–Neumann decomposes any epimorphism F2 → Cm*Cn so that the two free factors map onto the two cyclic factors, making every generating pair Nielsen-equivalent to the standard one. Theorem 2.4 is quoted correctly in the cases used. The only blemish is an apparent typo in the proof of Theorem 1.1(ii), where the entries of B are written in F_p instead of F_q; since s is subsequently taken in F_q and the computation is valid over F_q, this does not affect the argument.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper addresses the Neumann problem on lifting generating tuples along homomorphisms. For n=2, the authors exhibit infinitely many pairs (G,H), where G is a free product of finite cyclic groups and H=PSL(2,q), such that H is a homomorphic image of G but some generating pair of H is not the image of any generating pair of G under any homomorphism. Theorem 1.1 splits into four cases: G=C2*C3 with H=PSL(2,q) for q≥4 and q≠9; G=C2*C_p with H=PSL(2,p^k); G=C2*C_m with q≡3 mod4 under explicit arithmetic conditions on m; and G=C3*C3 with H=PSL(2,q) for q≥5. Corollary 1.2 produces the first negative examples in which H is infinite, for instance H=PSL(2,5)×G/G'' with G=PSL(2,Z). The method reduces the lifting question, via the Grushko–Neumann theorem, to whether a given Aut(F_2)-orbit of generating pairs of H contains an (m,n)-generating pair; a trace invariant of the commutator is then used to separate the desired orbit from all (m,n)-generating pairs.","tokens_in":11285,"tokens_out":16699,"duration_ms":152013,"significance":"If the main theorem is correct, this is a substantial advance on the Neumann problem: it replaces Dunwoody's single engineered soluble example with infinite families in which H is a simple projective special linear group, and it gives the first negative examples with H infinite. The method is transparent and reproducible: the key trace computations are explicit, the finite exceptional cases q=5,7 are checked by hand, and the appeal to external results (Grushko–Neumann, Dickson–Macbeath generation theorems, the McCullough–Wanderley classification of trace invariants, and Miller's commutator-order theorems) is clearly labelled. In particular, the load-bearing premise Fact 2.2 is a standard theorem rather than a circular assumption, and the reduction in Corollary 2.3 is sound. I found no load-bearing error in the central derivation.","major_comments":[],"minor_comments":[{"comment":"In the displayed matrix for B, the entries are declared to lie in F_p, but B is an element of SL(2,q) with q=p^k; the subsequent line puts s=b-c in F_q, so the entries should be in F_q. This is only a typo and does not affect the computation.","section":"§3, proof of Theorem 1.1(ii)"},{"comment":"The proof begins with 'Since A has order 4 in SL(2,q)', but the hypothesis is only A^4=I, which also allows A=-I of order 2. In the order-2 case the conclusion is trivial because then [A,B]=I and tr([A,B])=2, so the statement is correct; however, the proof as written should either restrict to the order-4 case used in Corollary 2.7 or mention the order-2 case separately.","section":"§2, Proposition 2.6"},{"comment":"The proof says only 'This follows directly from Corollary 2.7.' Since Corollary 2.7 assumes that H is (2,m)-generated, the proof should explicitly justify that the hypotheses on m imply this: if p divides m, use an element of order p; otherwise, a common divisor d≥3 of m with (q±1)/2 gives a possible element order d, and the Langer–Rosenberger theorem supplies (2,d)-generation, which implies (2,m)-generation.","section":"§3, proof of Theorem 1.1(iii)"}],"recommendation":"minor_revision","confidential_remarks":null},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Short version: this is a genuinely new supply of negative examples to the Neumanns' problem, and the proof approach works. Dunwoody already settled the yes/no question, but this paper shows the phenomenon is not rare: infinitely many pairs (C2*C3, PSL(2,q)), with extensions to C2*Cp, C2*Cm, C3*C3, and the first examples with H infinite. I went through the reduction and the trace computations; I did not find a load-bearing gap.\n\nThe main mechanism is clean. Corollary 2.3 reduces lifting to whether an A-orbit in Gamma_H contains an (m,n)-generating pair, using Nielsen transitivity on Gamma_{Cm*Cn}. That fact is the load-bearing external input, and it is correct via Grushko-Neumann/Lyndon. The trace invariant tau is constant on A-orbits, so any A-orbit whose trace value cannot occur for (m,n)-generating pairs is free. The hard finite checks in Lemmas 3.1 and 3.2 are a little compressed but correct; Theorem 2.4 is quoted accurately in the cases used. The small F_p instead of F_q typo in the proof of Theorem 1.1(ii) is presentation-level only; the computation is over F_q and the conclusion stands.\n\nThe paper is honest about where the method stops: q=9 is excluded, PSL(2,5) for C2*C5 is checked separately, and the paragraph after Theorem 1.1(iv) lists cases like (7,13), (19,37) where the trace invariant fills everything but 2. That limitation is stated rather than hidden, and it says more about the technique than about the theorem.\n\nSoft spots are minor: a few direct computations are left to the reader, the paper is not machine-checked, and the infinite-H construction via G/G'' is somewhat ancillary but works. None of this shakes the main result. The citation pattern is solid: Macbeath, Langer-Rosenberger, and McCullough-Wanderley are used as black boxes in exactly the ways one would expect.\n\nWho is this for: group theorists working on generation, free products, and PSL(2,q). A serious referee should engage with it. If I were the editor, I would send it out; the main theorems are very likely correct and worth having in the literature.","headline":"A solid construction paper: infinite families of free-product/PSL(2,q) negative examples for the Neumanns' problem, with a correct proof and only minor presentation gaps.","tokens_in":11839,"tokens_out":2643,"would_cite":true,"duration_ms":24529,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["20F05","20E06","20G40"],"pacs":[],"model":"deepseek-v4-flash","headline":"PSL(2,q) has generating pairs that cannot be lifted.","keywords":["generating tuples","lifting problem","free products","projective special linear groups","trace invariants","Nielsen transformations","modular group","PSL(2,q)"],"falsifier":"Choose $q=11$ and exhaustively list all generating pairs $(h_1,h_2)$ of $\\mathrm{PSL}(2,11)$ with $h_1^2=h_2^3=1$, computing $\\tau(h_1,h_2)=\\operatorname{tr}([h_1,h_2])$. The theorem predicts that no such pair has trace invariant $0$ and that none has commutator of order 2 or 3, since that would force the group they generate to be solvable; finding one would refute Theorem 1.1(i).","tokens_in":10963,"feed_emoji":"🧩","tokens_out":12675,"duration_ms":112602,"temperature":0.7,"pith_summary":"The paper addresses a classical question about generating tuples: if a group $G$ maps onto a group $H$, must every generating pair of $H$ be the image of some generating pair of $G$ along some homomorphism? It establishes that the answer is no for infinitely many natural pairs: take $G$ to be a free product of two finite cyclic groups, such as the modular group $C_2 * C_3$, and $H$ to be a finite projective special linear group $\\mathrm{PSL}(2,q)$. For all four families listed in Theorem 1.1, $H$ is a homomorphic image of $G$, yet some generating pair of $H$ cannot be obtained from any generating pair of $G$ along any homomorphism. A small modification yields the first negative examples in which $H$ is infinite, for instance $H = \\mathrm{PSL}(2,5) \\times G/G''$ with $G = C_2 * C_3$. The interest is that these are not exotic constructions: they are the classical modular group and its finite quotients.","feed_headline":"PSL(2,q) pairs that will not lift to the modular group","feed_subtitle":"A 1950s lifting problem has infinitely many negative answers, including the first with infinite target.","key_machinery":"The load-bearing object is the trace invariant. For a generating pair $(h_1,h_2)$ of $H=\\mathrm{PSL}(2,q)$, choose matrix representatives in $\\mathrm{SL}(2,q)$, form their commutator, and take its trace; the result $\\tau(h_1,h_2)\\in\\mathbb{F}_q$ is independent of the choice of representatives and is constant on the orbit of the pair under the automorphism group of the free group on two generators. Because generating pairs of a free product $C_m*C_n$ form a single such orbit, a pair of $H$ lifts to $G$ exactly when its orbit contains a pair $(h_1,h_2)$ with $h_1^m=h_2^n=1$. The proofs use a classification of which trace values occur for generating pairs of $\\mathrm{PSL}(2,q)$ together with classical facts about groups generated by two elements of prescribed orders whose commutator has order 2 or 3. The trace invariant is the bridge that turns a question about all homomorphisms into a finite check about one number.","core_discovery":"The central claim, stated as Theorem 1.1, is that the following pairs $(G,H)$ are negative examples to the lifting problem: (i) $G=C_2*C_3$, $H=\\mathrm{PSL}(2,q)$ for $q\\ge 4$, $q\\ne 9$; (ii) $G=C_2*C_p$, $H=\\mathrm{PSL}(2,q)$ with $q=p^k$, $p\\ge 3$, $q\\ge 7$, $q\\ne 9$; (iii) $G=C_2*C_m$, $H=\\mathrm{PSL}(2,q)$ with $q\\equiv 3 \\pmod 4$, $q\\ne 3$, and an additional divisibility condition linking $m$ to $p$, $(q+1)/2$, or $(q-1)/2$; and (iv) $G=C_3*C_3$, $H=\\mathrm{PSL}(2,q)$ for $q\\ge 5$. In each case $H$ is a homomorphic image of $G$, but there exist generating pairs $(h_1,h_2)$ of $H$ such that for every homomorphism $\\vartheta\\colon G\\to H$ and every generating pair $(g_1,g_2)$ of $G$, the images $(g_1^\\vartheta,g_2^\\vartheta)$ are not equal to $(h_1,h_2)$. Corollary 1.2 transfers the phenomenon to infinite $H$ by taking a direct product with the infinite group $G/G''$; the projection back to the finite factor would turn any lift of the product pair into a lift of the original non-liftable pair.","pith_inferences":["One could extend the same orbit-and-trace test to other non-solvable quotients of free products of cyclic groups: whenever the set of trace invariants is known, the existence of non-liftable pairs reduces to a finite computation, potentially producing negative examples outside the $\\mathrm{PSL}(2,q)$ family.","The residue condition $q\\equiv 3 \\pmod 4$ in part (iii) is likely an artifact of the proof; the authors note small cases where the invariant takes all but one value, so a refined invariant or a different choice of orbit might remove the condition.","The general recipe behind the proofs suggests that any orbit whose commutator has an order forcing solvability of two-generator quotients with the prescribed element orders will give a non-lifting example in a non-solvable target, a criterion that could be tested on other finite non-solvable groups.","For $n\\ge 3$, no negative examples are known; since generating $n$-tuples of a free product do not form a single orbit under $\\mathrm{Aut}(F_n)$, new ideas are needed, but partial results might come from studying orbits of $n$-tuples with fixed trace-like invariants."],"forward_implications":["For every prime $p\\ge 5$, the quotient $\\mathrm{PSL}(2,p)$ of the modular group $C_2*C_3$ has generating pairs that no homomorphism from the modular group can lift; the finite-quotient picture of the modular group is therefore not tuple-surjective.","Non-lifting is not a pathology of one crafted pair of solvable groups: it occurs for infinitely many pairs in which $G$ is a free product of cyclic groups and $H$ is one of the most studied finite simple groups.","The first negative examples with infinite $H$ follow formally from the finite ones: $H=\\mathrm{PSL}(2,5)\\times G/G''$ is a quotient of $G=C_2*C_3$ but has non-liftable generating pairs.","The obstruction is visible in a single scalar invariant, so the same orbit-and-trace test can be run for any finite quotient for which the trace values on generating pairs are known."],"supporting_citations":[{"why":"poses the lifting problem for generating tuples that the paper answers in the negative.","marker":"[14]"},{"why":"supplies the first negative example, the construction the present paper contrasts with.","marker":"[3]"},{"why":"proves the positive result for epimorphisms with finite kernel, the boundary that the new examples push against.","marker":"[4]"},{"why":"contains the classical fact that all generating pairs of a free product of cyclic groups form one orbit under the automorphism group of the free group on two generators.","marker":"[16]"},{"why":"classifies when $\\mathrm{PSL}(2,q)$ is generated by elements of prescribed orders, supporting the claim that $H$ is a homomorphic image of $G$.","marker":"[10]"},{"why":"provides Theorem 2.4, the classification of trace invariants of generating pairs used to find orbits with forbidden values.","marker":"[11]"},{"why":"supplies the classical results on groups generated by elements of order 2 and 3 with commutator of order 2 or 3, used to show that certain orbits are (2,3)-free.","marker":"[12]"},{"why":"records the observation that the set of commutators in an automorphism-group orbit lies in two conjugacy classes, giving the trace invariant its constancy.","marker":"[13]"}],"fun_headline_variants":["Infinitely many free product-PSL(2,q) pairs fail to lift","Lifting problem fails for infinitely many C2*C3 to PSL(2,q)","First infinite-target counterexamples to Neumann's lifting problem","No lift: free products to PSL(2,q) for infinitely many q","Neumann's lifting problem: free products to PSL(2,q) fail"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The argument rests on the fact that all generating pairs of a free product $C_m*C_n$ form a single orbit under the automorphism group of the free group on two generators; if that transitivity failed, a pair with a forbidden trace invariant could still lift through a pair from a different orbit.","fun_headline_variants_meta":{"raw":{"variants":["Infinitely many free product-PSL(2,q) pairs fail to lift","Lifting problem fails for infinitely many C2*C3 to PSL(2,q)","First infinite-target counterexamples to Neumann's lifting problem","No lift: free products to PSL(2,q) for infinitely many q","Neumann's lifting problem: free products to PSL(2,q) fail"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.001647,"raw_usage":{"total_tokens":6653,"prompt_tokens":1164,"completion_tokens":5489,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":780,"completion_tokens_details":{"reasoning_tokens":5388}},"tokens_in":780,"tokens_out":5489,"duration_ms":40254,"temperature":1.0,"reasoning_tokens":5388,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T13:15:04.491547+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Choose $q=11$ and exhaustively list all generating pairs $(h_1,h_2)$ of $\\mathrm{PSL}(2,11)$ with $h_1^2=h_2^3=1$, computing $\\tau(h_1,h_2)=\\operatorname{tr}([h_1,h_2])$. The theorem predicts that no such pair has trace invariant $0$ and that none has commutator of order 2 or 3, since that would force the group they generate to be solvable; finding one would refute Theorem 1.1(i).","supporting_citations":[{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"poses the lifting problem for generating tuples that the paper answers in the negative."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"supplies the first negative example, the construction the present paper contrasts with."},{"cited_title":"Gasch¨ utz, Zu einem von B","cited_arxiv_id":null,"evidence_quote":"proves the positive result for epimorphisms with finite kernel, the boundary that the new examples push against."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"contains the classical fact that all generating pairs of a free product of cyclic groups form one orbit under the automorphism group of the free group on two generators."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"classifies when $\\mathrm{PSL}(2,q)$ is generated by elements of prescribed orders, supporting the claim that $H$ is a homomorphic image of $G$."},{"cited_title":"McCullough and M","cited_arxiv_id":null,"evidence_quote":"provides Theorem 2.4, the classification of trace invariants of generating pairs used to find orbits with forbidden values."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"supplies the classical results on groups generated by elements of order 2 and 3 with commutator of order 2 or 3, used to show that certain orbits are (2,3)-free."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"records the observation that the set of commutators in an automorphism-group orbit lies in two conjugacy classes, giving the trace invariant its constancy."}],"review_version":1}